$8$ people including $A,B,C$, and $D$ will be rearranged. In how many ways can they be rearranged such that $B$ and $C$ will be between $A$ and $D$? 
$8$ people including $A,B,C$, and $D$ will be rearranged. In how many ways can they be rearranged such that $B$ and $C$ will be between $A$ and $D$?

I tried to break this into cases. However, it didn't work. 
Regards
 A: First choose 4 places, that you can do on ${8\choose 4}$ ways, then on edge put $A$ and $D$, you can do that on 2 ways and in the midle $B$ and $C$, again on 2 ways. Then arrange all others on remaining places, that is on 4! ways. Now multiply all these. So the answer is $${8\choose 4}\cdot 2\cdot 2\cdot 4! = 8!/6$$
A: We first choose among the $8$ positions the $4$ ones for $A,B,C,D$. This can be done in $\binom{8}{4}$ ways. These positions cab be filled in $4$ ways: $ABCD$, $ACBD$, $DBCA$, $DCBA$. The remanining $8-4$ positions can be filled in $4!$  ways. Hence the total number of arrangements is
$${8\choose 4}\cdot 4 \cdot 4!$$
A: This answer assumes that ABCD is unique from ACBD, though I can answer for the other case as well if that is the case.
N = number of spots (8 in your case)
2 * summation(n = 1, N - 3){n! * (N+1-n)!}
In your case, this simplifies to 
Take the two boundary tokens and orient them on the edges. There are two "symmetric" orientations for each orientation of the boundary tokens (A******D is distinct from D******A, and there are only two orientations of two tokens with two spots.) From there, you have the inner spaces, designated as "Sep". For each value Sep, there are Sep! permutations inside the boundary tokens, and (N - (Sep + 2))! permutations outside the boundary, so for each value Sep, there are Sep! * (N - (Sep + 2))! * 2 * (number of "slide" positions). The "Slide" positions are the orientations you can bring A and D to through sliding; in simple terms, if the string was A****D**, Slide is 3, since you can have A****D**, *A****D*, **A****D. Factor the 2 out and do some refactoring, plug in N = 8, and you get 2976.
EDIT: My final sum of 2976 was a quick desmos plugin, so if anyone would like to plug in the equation I provided above to verify, you can go ahead and do that.
